Math Core

Determinants give you exact formulas where row reduction gives only an algorithm: a formula for each variable in a system $A\mathbf{x} = \mathbf{b}$ , a formula for every entry of $A^{-1}$ , and a formula for the area or volume of a parallelogram or parallelepiped. The last one explains the claim from the start of the unit that $\det A$ is the factor by which $A$ scales area.

Cramer's rule

For an $n \times n$ matrix $A$ and a vector $\mathbf{b}$ in $\mathbb{R}^n$ , write $A_i(\mathbf{b})$ for the matrix you get by replacing column $i$ of $A$ with $\mathbf{b}$ .

Cramer's rule

If $A$ is an invertible $n \times n$ matrix, then for every $\mathbf{b}$ in $\mathbb{R}^n$ the unique solution of $A\mathbf{x} = \mathbf{b}$ has entries

x_i = \frac{\det A_i(\mathbf{b})}{\det A}, \qquad i = 1, 2, \ldots, n.

The proof is short and worth seeing. Let $I_i(\mathbf{x})$ be the identity matrix with column $i$ replaced by $\mathbf{x}$ . Multiplying by $A$ acts column by column: $A\mathbf{e}_j = \mathbf{a}_j$ for $j \neq i$ , and $A\mathbf{x} = \mathbf{b}$ in column $i$ . So

A \, I_i(\mathbf{x}) = A_i(\mathbf{b}).

Expanding $\det I_i(\mathbf{x})$ across row $i$ shows it equals $x_i$ . Taking determinants and using the multiplicative property, $(\det A)\,x_i = \det A_i(\mathbf{b})$ . Divide by $\det A \neq 0$ and you're done.

Worked example: Cramer's rule for a 2 × 2 system

Solve $\begin{aligned} 3x_1 - 2x_2 &= 11 \\ 4x_1 + 5x_2 &= 7 \end{aligned}$ .

\det A = \begin{vmatrix} 3 & -2 \\ 4 & 5 \end{vmatrix} = 15 + 8 = 23.

Replace column 1, then column 2, by $\mathbf{b} = (11, 7)$ :

\det A_1(\mathbf{b}) = \begin{vmatrix} 11 & -2 \\ 7 & 5 \end{vmatrix} = 55 + 14 = 69, \qquad \det A_2(\mathbf{b}) = \begin{vmatrix} 3 & 11 \\ 4 & 7 \end{vmatrix} = 21 - 44 = -23.

So $x_1 = \dfrac{69}{23} = 3$ and $x_2 = \dfrac{-23}{23} = -1$ . Check: $3(3) - 2(-1) = 11$ and $4(3) + 5(-1) = 7$ . The solution is $(3, -1)$ .

Worked example: Cramer's rule for a 3 × 3 system

Solve $\begin{aligned} x_1 + 2x_2 + x_3 &= 4 \\ x_2 + 3x_3 &= -1 \\ 2x_1 - x_2 + x_3 &= -1 \end{aligned}$ .

Expanding across row 1 each time:

\begin{aligned} \det A &= \begin{vmatrix} 1 & 2 & 1 \\ 0 & 1 & 3 \\ 2 & -1 & 1 \end{vmatrix} = 1(1 + 3) - 2(0 - 6) + 1(0 - 2) = 14, \\ \det A_1(\mathbf{b}) &= \begin{vmatrix} 4 & 2 & 1 \\ -1 & 1 & 3 \\ -1 & -1 & 1 \end{vmatrix} = 4(4) - 2(2) + 1(2) = 14, \\ \det A_2(\mathbf{b}) &= \begin{vmatrix} 1 & 4 & 1 \\ 0 & -1 & 3 \\ 2 & -1 & 1 \end{vmatrix} = 1(2) - 4(-6) + 1(2) = 28, \\ \det A_3(\mathbf{b}) &= \begin{vmatrix} 1 & 2 & 4 \\ 0 & 1 & -1 \\ 2 & -1 & -1 \end{vmatrix} = 1(-2) - 2(2) + 4(-2) = -14. \end{aligned}

So $x_1 = \dfrac{14}{14} = 1$ , $x_2 = \dfrac{28}{14} = 2$ , $x_3 = \dfrac{-14}{14} = -1$ , and the solution is $(1, 2, -1)$ .

Common mistake

Cramer's rule applies only when $\det A \neq 0$ . If $\det A = 0$ , the system has either no solution or infinitely many, and Cramer's rule says nothing about which; use row reduction. Also, for anything bigger than $3 \times 3$ , Cramer's rule is far slower than row reduction. Its value is theoretical: it shows exactly how the solution depends on the entries of $A$ and $\mathbf{b}$ .

A formula for the inverse

Column $j$ of $A^{-1}$ is the solution of $A\mathbf{x} = \mathbf{e}_j$ . Apply Cramer's rule to that system and expand $\det A_i(\mathbf{e}_j)$ down column $i$ : the only nonzero entry is the $1$ in row $j$ , which leaves the cofactor $C_{ji}$ . So the $(i, j)$ entry of $A^{-1}$ is $C_{ji}/\det A$ .

Definition

Adjugate

The adjugate of $A$ , written $\operatorname{adj} A$ , is the transpose of the matrix of cofactors: its $(i, j)$ entry is $C_{ji}$ . If $A$ is invertible,

A^{-1} = \frac{1}{\det A}\operatorname{adj} A.

Notice the index swap: the $(i, j)$ entry of the inverse uses the cofactor from position $(j, i)$ . For a $2 \times 2$ matrix this formula is the familiar $\dfrac{1}{ad - bc}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$ . For larger matrices it's mainly useful when you need just one entry of $A^{-1}$ .

Area of a parallelogram

Area and volume

If $A$ is $2 \times 2$ , the area of the parallelogram spanned by the columns of $A$ is $|\det A|$ .

If $A$ is $3 \times 3$ , the volume of the parallelepiped spanned by the columns of $A$ is $|\det A|$ .

Here is why. For a diagonal matrix $\begin{bmatrix} a & 0 \\ 0 & d \end{bmatrix}$ the parallelogram is a rectangle with area $|a||d| = |\det A|$ . Any other matrix can be reduced to diagonal form by column interchanges and column replacements. An interchange just relabels the two edges (same parallelogram, sign of the determinant flips, absolute value unchanged). A replacement $\mathbf{a}_1 \to \mathbf{a}_1 + c\,\mathbf{a}_2$ slides one edge along a line parallel to the other: that's a shear, which keeps the same base and height, so the area stays the same, and so does the determinant. The same argument works in three dimensions with base area times height.

Because the determinant doesn't care whether you use rows or columns, you can put the vectors in as rows or as columns.

Worked example: A parallelogram not at the origin

Find the area of the parallelogram with vertices $(-1, 1)$ , $(3, 2)$ , $(0, 4)$ and $(4, 5)$ .

First translate so a vertex sits at the origin; translation doesn't change area. The two edges leaving $(-1, 1)$ are

(3, 2) - (-1, 1) = (4, 1), \qquad (0, 4) - (-1, 1) = (1, 3).

(Check: $(-1, 1) + (4, 1) + (1, 3) = (4, 5)$ , the fourth vertex.)

The parallelogram with edges (4, 1) and (1, 3) leaving the vertex (-1, 1).Open in grapher →

\text{Area} = \left|\begin{vmatrix} 4 & 1 \\ 1 & 3 \end{vmatrix}\right| = |12 - 1| = 11.

A triangle with the same three vertices $(-1, 1)$ , $(3, 2)$ , $(0, 4)$ is half of this parallelogram, so its area is $\dfrac{11}{2}$ .

In three dimensions it works the same way: the parallelepiped spanned by $(2, 0, 1)$ , $(1, 3, 0)$ and $(0, 1, 4)$ has volume

\left|\begin{vmatrix} 2 & 1 & 0 \\ 0 & 3 & 1 \\ 1 & 0 & 4 \end{vmatrix}\right| = |2(12 - 0) - 1(0 - 1) + 0| = 25.

Linear transformations scale area by $|\det A|$

Now the big picture. Let $T(\mathbf{x}) = A\mathbf{x}$ with $A$ a $2 \times 2$ matrix. A parallelogram spanned by $\mathbf{p}$ and $\mathbf{q}$ is carried to the parallelogram spanned by $A\mathbf{p}$ and $A\mathbf{q}$ , whose area is

\bigl|\det[\,A\mathbf{p} \;\; A\mathbf{q}\,]\bigr| = \bigl|\det(A[\,\mathbf{p} \;\; \mathbf{q}\,])\bigr| = |\det A| \cdot \bigl|\det[\,\mathbf{p} \;\; \mathbf{q}\,]\bigr|.

So every parallelogram's area is multiplied by the same factor $|\det A|$ . Any reasonable region can be approximated by tiny squares, so the same holds for all of them.

Scaling by a linear transformation

If $T(\mathbf{x}) = A\mathbf{x}$ with $A$ a $2 \times 2$ matrix and $S$ is a region of finite area in $\mathbb{R}^2$ , then

\text{area of } T(S) = |\det A| \cdot \text{area of } S.

The same holds for volume in $\mathbb{R}^3$ with $A$ a $3 \times 3$ matrix.

Worked example: The area of an ellipse

Find the area enclosed by the ellipse $\dfrac{x^2}{9} + \dfrac{y^2}{4} = 1$ .

The matrix $A = \begin{bmatrix} 3 & 0 \\ 0 & 2 \end{bmatrix}$ sends the unit disk $u^2 + v^2 \le 1$ to this ellipse region: if $(x, y) = (3u, 2v)$ , then $\dfrac{x^2}{9} + \dfrac{y^2}{4} = u^2 + v^2$ . The unit disk has area $\pi$ , and $\det A = 6$ , so the ellipse encloses area $6\pi$ . In general, the ellipse with semi-axes $a$ and $b$ has area $\pi ab$ .

Tip

A negative determinant is not a mistake. Its sign tells you the transformation reverses orientation (like a reflection). Area is always the absolute value.

Practice

Practice 1

Use Cramer's rule to solve $\begin{aligned} 3x_1 + 2x_2 &= 7 \\ 5x_1 + 4x_2 &= 13 \end{aligned}$ .

Enter a point like (2, -3)

Practice 2

Use Cramer's rule to solve $\begin{aligned} 2x_1 - x_2 &= 4 \\ 3x_1 + 4x_2 &= 1 \end{aligned}$ .

Enter a point like (2, -3)

Practice 3

Use Cramer's rule to find only $x_2$ for the system

\begin{aligned} 2x_1 + x_2 - x_3 &= 3 \\ x_1 + 3x_2 + 2x_3 &= 1 \\ x_1 + x_3 &= 4 \end{aligned}